Practicing Success
The set of all x values for which |f(x) + g(x)| < |f(x)| + |g(x)| is true if f(x) = x – 3 and g(x) = 4 – x is given by |
R R -]3, 4[ R –[3, 4] None of these |
R –[3, 4] |
$|f(x) + g(x)| < |f(x)| + |g(x)|$ $⇒|x-3+4-x| < |x-3|+|4-x|$ $1<|x-3|+|4-x|$ for equation if $x∈(3,4)$ then inequality doesn't hold so R –[3, 4] |